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4.5The equation of state of an ideal elastic substance is T=k theta((L)/(L_(0))-(L_(0)^(2))/(L^(2))) where 7 is the tension and theta the temperature.k is a constant and L_(0) (the value of L al tension) is a function of temperature only. Show that the isothermal Young's modulus is given by Y=(k theta)/(A)((L)/(L_(0)) (2L_(0)^(2))/(L^(2))) and that its value at zero tension is given Y_(0)=(3k theta)/(A) where A is the cross-sectional ara of the wire.?
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4.5The equation of state of an ideal elastic substance is T=k theta((L...
Equation of State
The equation of state of an ideal elastic substance is given by:
T = kθ((L/L_0) - (L_0^2/L^2))
where T is the tension, θ is the temperature, k is a constant, L_0 is the value of L at zero tension (which is a function of temperature only), and L is the current length of the substance.
Deriving the Isothermal Young's Modulus
The Young's modulus (Y) is a measure of the stiffness of a material. It is defined as the ratio of stress to strain. In this case, since the substance is ideal and elastic, we can consider it to be under isothermal conditions.
The stress (σ) is given by the tension divided by the cross-sectional area (A) of the substance:
σ = T/A
The strain (ε) is the change in length (ΔL) divided by the original length (L):
ε = ΔL/L
Step 1: Expressing Stress in terms of L and L_0
From the equation of state, we can express the tension in terms of L and L_0:
T = kθ((L/L_0) - (L_0^2/L^2))
Substituting this value of T into the stress equation:
σ = (kθ((L/L_0) - (L_0^2/L^2)))/A
Step 2: Expressing Strain in terms of L and L_0
The change in length (ΔL) can be expressed as the difference between the current length (L) and the original length (L_0):
ΔL = L - L_0
Substituting this value of ΔL into the strain equation:
ε = (L - L_0)/L
Step 3: Deriving the Isothermal Young's Modulus
Now we can substitute the expressions for stress and strain into the definition of Young's modulus:
Y = σ/ε
Substituting the expressions for stress and strain:
Y = [(kθ((L/L_0) - (L_0^2/L^2)))/A]/[(L - L_0)/L]
Simplifying the expression:
Y = (kθ/L_0)(L/(L - L_0)) * (L^2/[(L - L_0)A])
Y = (kθ/L_0)(L^2/(L - L_0)A)
Step 4: Young's Modulus at Zero Tension
At zero tension, the length of the substance is equal to the original length:
L = L_0
Substituting this value into the expression for Young's modulus:
Y_0 = (kθ/L_0)(L_0^2/(L_0 - L_0)A)
Simplifying the expression:
Y_0 = (kθ/L_0)(L_0^2/0A)
Since the denominator is zero, the Young's modulus at zero tension is undefined.
However, if we take the limit as T approaches zero, we can find the value of Young's modulus:
lim(T→0) Y = (kθ/L_0)(L_
The equation of state of an ideal elastic substance is given by:
T = kθ((L/L_0) - (L_0^2/L^2))
where T is the tension, θ is the temperature, k is a constant, L_0 is the value of L at zero tension (which is a function of temperature only), and L is the current length of the substance.
Deriving the Isothermal Young's Modulus
The Young's modulus (Y) is a measure of the stiffness of a material. It is defined as the ratio of stress to strain. In this case, since the substance is ideal and elastic, we can consider it to be under isothermal conditions.
The stress (σ) is given by the tension divided by the cross-sectional area (A) of the substance:
σ = T/A
The strain (ε) is the change in length (ΔL) divided by the original length (L):
ε = ΔL/L
Step 1: Expressing Stress in terms of L and L_0
From the equation of state, we can express the tension in terms of L and L_0:
T = kθ((L/L_0) - (L_0^2/L^2))
Substituting this value of T into the stress equation:
σ = (kθ((L/L_0) - (L_0^2/L^2)))/A
Step 2: Expressing Strain in terms of L and L_0
The change in length (ΔL) can be expressed as the difference between the current length (L) and the original length (L_0):
ΔL = L - L_0
Substituting this value of ΔL into the strain equation:
ε = (L - L_0)/L
Step 3: Deriving the Isothermal Young's Modulus
Now we can substitute the expressions for stress and strain into the definition of Young's modulus:
Y = σ/ε
Substituting the expressions for stress and strain:
Y = [(kθ((L/L_0) - (L_0^2/L^2)))/A]/[(L - L_0)/L]
Simplifying the expression:
Y = (kθ/L_0)(L/(L - L_0)) * (L^2/[(L - L_0)A])
Y = (kθ/L_0)(L^2/(L - L_0)A)
Step 4: Young's Modulus at Zero Tension
At zero tension, the length of the substance is equal to the original length:
L = L_0
Substituting this value into the expression for Young's modulus:
Y_0 = (kθ/L_0)(L_0^2/(L_0 - L_0)A)
Simplifying the expression:
Y_0 = (kθ/L_0)(L_0^2/0A)
Since the denominator is zero, the Young's modulus at zero tension is undefined.
However, if we take the limit as T approaches zero, we can find the value of Young's modulus:
lim(T→0) Y = (kθ/L_0)(L_
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4.5The equation of state of an ideal elastic substance is T=k theta((L...
Be a theta the temperature. and al tension is a function that is a cross sectional
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4.5The equation of state of an ideal elastic substance is T=k theta((L)/(L_(0))-(L_(0)^(2))/(L^(2))) where 7 is the tension and theta the temperature.k is a constant and L_(0) (the value of L al tension) is a function of temperature only. Show that the isothermal Young's modulus is given by Y=(k theta)/(A)((L)/(L_(0)) (2L_(0)^(2))/(L^(2))) and that its value at zero tension is given Y_(0)=(3k theta)/(A) where A is the cross-sectional ara of the wire.?
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4.5The equation of state of an ideal elastic substance is T=k theta((L)/(L_(0))-(L_(0)^(2))/(L^(2))) where 7 is the tension and theta the temperature.k is a constant and L_(0) (the value of L al tension) is a function of temperature only. Show that the isothermal Young's modulus is given by Y=(k theta)/(A)((L)/(L_(0)) (2L_(0)^(2))/(L^(2))) and that its value at zero tension is given Y_(0)=(3k theta)/(A) where A is the cross-sectional ara of the wire.? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about 4.5The equation of state of an ideal elastic substance is T=k theta((L)/(L_(0))-(L_(0)^(2))/(L^(2))) where 7 is the tension and theta the temperature.k is a constant and L_(0) (the value of L al tension) is a function of temperature only. Show that the isothermal Young's modulus is given by Y=(k theta)/(A)((L)/(L_(0)) (2L_(0)^(2))/(L^(2))) and that its value at zero tension is given Y_(0)=(3k theta)/(A) where A is the cross-sectional ara of the wire.? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 4.5The equation of state of an ideal elastic substance is T=k theta((L)/(L_(0))-(L_(0)^(2))/(L^(2))) where 7 is the tension and theta the temperature.k is a constant and L_(0) (the value of L al tension) is a function of temperature only. Show that the isothermal Young's modulus is given by Y=(k theta)/(A)((L)/(L_(0)) (2L_(0)^(2))/(L^(2))) and that its value at zero tension is given Y_(0)=(3k theta)/(A) where A is the cross-sectional ara of the wire.?.
4.5The equation of state of an ideal elastic substance is T=k theta((L)/(L_(0))-(L_(0)^(2))/(L^(2))) where 7 is the tension and theta the temperature.k is a constant and L_(0) (the value of L al tension) is a function of temperature only. Show that the isothermal Young's modulus is given by Y=(k theta)/(A)((L)/(L_(0)) (2L_(0)^(2))/(L^(2))) and that its value at zero tension is given Y_(0)=(3k theta)/(A) where A is the cross-sectional ara of the wire.? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about 4.5The equation of state of an ideal elastic substance is T=k theta((L)/(L_(0))-(L_(0)^(2))/(L^(2))) where 7 is the tension and theta the temperature.k is a constant and L_(0) (the value of L al tension) is a function of temperature only. Show that the isothermal Young's modulus is given by Y=(k theta)/(A)((L)/(L_(0)) (2L_(0)^(2))/(L^(2))) and that its value at zero tension is given Y_(0)=(3k theta)/(A) where A is the cross-sectional ara of the wire.? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 4.5The equation of state of an ideal elastic substance is T=k theta((L)/(L_(0))-(L_(0)^(2))/(L^(2))) where 7 is the tension and theta the temperature.k is a constant and L_(0) (the value of L al tension) is a function of temperature only. Show that the isothermal Young's modulus is given by Y=(k theta)/(A)((L)/(L_(0)) (2L_(0)^(2))/(L^(2))) and that its value at zero tension is given Y_(0)=(3k theta)/(A) where A is the cross-sectional ara of the wire.?.
Solutions for 4.5The equation of state of an ideal elastic substance is T=k theta((L)/(L_(0))-(L_(0)^(2))/(L^(2))) where 7 is the tension and theta the temperature.k is a constant and L_(0) (the value of L al tension) is a function of temperature only. Show that the isothermal Young's modulus is given by Y=(k theta)/(A)((L)/(L_(0)) (2L_(0)^(2))/(L^(2))) and that its value at zero tension is given Y_(0)=(3k theta)/(A) where A is the cross-sectional ara of the wire.? in English & in Hindi are available as part of our courses for Physics.
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ample number of questions to practice 4.5The equation of state of an ideal elastic substance is T=k theta((L)/(L_(0))-(L_(0)^(2))/(L^(2))) where 7 is the tension and theta the temperature.k is a constant and L_(0) (the value of L al tension) is a function of temperature only. Show that the isothermal Young's modulus is given by Y=(k theta)/(A)((L)/(L_(0)) (2L_(0)^(2))/(L^(2))) and that its value at zero tension is given Y_(0)=(3k theta)/(A) where A is the cross-sectional ara of the wire.? tests, examples and also practice Physics tests.
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